If $a, b, c$ are lengths of the sides $BC, CA, AB$ respectively of $\triangle ABC$ and $H$ is any point in the plane of $\triangle ABC$ such that $a \vec{AH} + b \vec{BH} + c \vec{CH} = \vec{0}$,then $H$ is the

The projection of $\overline{AB}$ on $\overline{CD}$,where $A \equiv (2, -3, 0)$,$B \equiv (1, -4, -2)$,$C \equiv (4, 6, 8)$,and $D \equiv (7, 0, 10)$ is

If $a, b$ and $c$ are three vectors such that $|a + b + c| = 1$, $c = \lambda(a \times b)$, and $|a| = \frac{1}{\sqrt{3}}$, $|b| = \frac{1}{\sqrt{2}}$, $|c| = \frac{1}{\sqrt{6}}$, then the angle between $a$ and $b$ is

The projection of the vector $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$ on the vector $\vec{b} = 4\hat{i} - 4\hat{j} + 7\hat{k}$ is:

$\vec{a}, \vec{b}, \text{ and } \vec{c}$ are three vectors such that $|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$. If $\vec{a}, \vec{b}, \vec{c}$ are perpendicular to the vectors $\vec{b}+\vec{c}, \vec{c}+\vec{a}, \vec{a}+\vec{b}$ respectively,then $\sqrt{|\vec{a}+\vec{b}+\vec{c}|^2-2} = $

Two adjacent sides of a parallelogram are represented by $\vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side $\vec{AD}$ is rotated by an acute angle $\theta$ in the plane of the parallelogram such that $\vec{AD}$ becomes $\vec{AD'}$. If $\vec{AD'}$ is perpendicular to $\vec{AB}$,find $\cos \theta$.